91Ó°ÊÓ

Trigonometric substitution is a powerful technique in calculus that simplifies complex expressions by substituting trigonometric functions for variables. In this exercise, we are using the substitution \( x = \cos \theta \). This allows us to exploit the well-known trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \). In doing this, we can express \( \sin \theta \) in terms of \( x \), which results in \( \sin \theta = \sqrt{1 - x^2} \).

By making these transformations, we are preparing the ground for easier differentiation and manipulation of expressions. With trigonometric substitution, we can convert complicated parts of a differential equation into simpler forms, often reducing algebraic complexity and making the calculus steps more straightforward.

This approach is particularly useful when dealing with integrals and differential equations that contain square roots of quadratic expressions. We can thereby translate problems into a solvable form by leveraging the power of trigonometric identities.

By applying trigonometric substitutions, we transform the second derivative from \( \frac{d^2 y}{d\theta^2} \) to a function in terms of \( x \). The transformation is based on the result \( \frac{d^2 y}{d \theta^2} = \sin^2 \theta \frac{d^2 y}{dx^2} - \cos \theta \frac{dy}{dx} \).

Here, the identities \( \sin \theta = \sqrt{1-x^2} \) and \( \cos \theta = x \) help express the second derivative using \( x \) instead of \( \theta \). This change redeems a complex trigonometric form into an algebraic one.

What this transformation effectively does is convert every instance of something defined over \( \theta \) to a similar expression defined in the simpler terms of \( x \). Making such transformations easier to work with is a common theme in solving differential equations where simplification and pattern recognition help reach the solution.

Mathematical proof is the logical process used to demonstrate the truth of a statement or equation. In this exercise, we need to prove that if \( x = \cos \theta \), then the given complex trigonometric equation transforms into a simpler differential equation.

We start with the transformed expressions for derivatives and substitute them into the longer expression \( \sin \theta \frac{d^2 y}{d\theta^2} + \cos \theta \frac{dy}{d\theta} + n(n+1)(\sin \theta) y = 0 \). The aim here is to show that this equation holds true by simplifying each side and demonstrating equality.

This proof then confirms that the requisite transformation steps bridge the complex domain of \( \theta \) with the algebraic simplicity regarding \( x \), confirming the initial hypothesis of the transformation being valid. Thus, rigorous proof exercises the logical structure and precision required in mathematics to verify any derived conclusion.

Problem solving in mathematics involves tackling a sequence of logical and sometimes complex steps. Here is how we approached this exercise:

• First, we understood the statement of the problem, identifying how substituting \( x = \cos \theta \) forms the foundation.
• Next, we transformed the first derivative \( \frac{dy}{d\theta} \) using the trigonometric identity and substitution, simplifying it with \( \sin \theta = \sqrt{1-x^2} \).
• The second derivative was similarly transformed, expressing both derivatives in terms of \( x \) using the relationships from trigonometric identities.
• After that, we applied the derived expressions to the given equation, aiming to find our final target differential equation.
• Finally, by simplifying and verifying each step, we matched the terms with our goal equation \( (1-x^2) \frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} + n(n+1)y = 0 \).

These structured problem solving steps ensure that each piece of the puzzle fits correctly, building confidence in the equation's validity and providing a clear pathway to the solution.